This section covers:

**Introduction to Imaginary Numbers****Working with “\(i\)***”***Quadratic Formula with Complex Solutions****Completing the Square with Complex Solutions****More Practice**

# Introduction to Imaginary Numbers

Think of **imaginary numbers** as numbers that are typically used in mathematical computations to get to/from “real” numbers (because they are more easily used in advanced computations), but really don’t exist in life as we know it. Yet they are real in the sense that they do exist and can be explained quite easily in terms of math as the **square root of a negative number**.

Actually, imaginary numbers are used quite frequently in engineering and physics, such as an alternating current in electrical engineering, which is usually represented by a complex number. (Don’t worry; I don’t know too much about alternating currents either.)

Let’s look again at the Venn Diagram from the **Types of Numbers and Algebraic Properties** section:

A **complex number** consists of a “real” part and an “imaginary” (non-real) part, and typically looks like \(a+bi\), where “\(a\)” is the real part, and “\(b\)” is the imaginary part, following by “\(i\)”, to indicate the “imaginary” unit. Note that complex numbers consist of both real numbers (\(a+0i\), such as **3**) and non-real numbers (\(a+bi,\,\,\,b\ne 0\), such as \(3+i\)); thus, all real numbers are also complex.

An imaginary number is the “\(i\)” part of a real number, and exists when we have to take the square root of a negative number. So technically, an imaginary number is only the “\(i\)” part of a complex number, and a pure imaginary number is a complex number that has no real part. It can get a little confusing!

Now let’s look at the following graph and notice that the parabola **never** touches the \(x\)-axis, so there aren’t any \(x\)-intercepts, although the “roots”, “zeros”, “solutions”, and “values” are “complex” or “imaginary”.

As we learned in the **Introduction to Quadratics** section, the **discriminant** can be used to determine what type of solutions a quadratic has. For imaginary solutions, since the graph has no roots, it has a discriminant \(\left( {{{b}^{2}}-4ac} \right)\) that is less than **0**. The graph will never touch the \(x\)-axis, yet we can still find imaginary roots, and the roots will have “\(i\)”s in them, as we see later. But we can’t find these roots with a graphing calculator!

# Working with “\(i\)”, the \(\sqrt{{-1}}\)

Before working problems that have imaginary solutions, we need to learn about the value of a special “number” called “\(i\)”. \(i\)** **is simply \(\sqrt{{-1}}\), which can’t exist in our “real” system, since we can never take two “real” numbers multiplied together to get \(–1\). Since \(i\) equals \(\sqrt{{-1}}\), then it follows that:

## \({{i}^{{\,2}}}\,\,=\,\,\,-1\)

So, \(\displaystyle \color{#800000}{{\sqrt{{-16}}}}=4i\), since \(\displaystyle \sqrt{{-16}}=\sqrt{{\left( {16} \right)\times \left( {-1} \right)}}=\left( {\sqrt{{16}}} \right)\left( {\sqrt{{-1}}} \right)=4i\). Similarly, \(\color{#800000}{{\sqrt{{-3}}}}=i\sqrt{3}\), or \(\sqrt{3}\,i\) (if you put the \(i\) at the end, make sure it is clearly outside of the square root sign in this case).

And here’s something really cool: when we multiply *i*’s together, we notice a pattern:

Note that there’s a repeating pattern when raising “*i*” to an exponent. Notice that every **fourth **exponent number repeats, so \({{i}^{1}}={{i}^{5}}={{i}^{9}}=i\), and so on. Because of this, we can easily compute “\(i\)” raised to any exponent by dividing that exponent by **four**, and examining the remainder, as shown in the examples:

\(\displaystyle \begin{array}{l}\color{#800000}{{{{i}^{{77}}}}}=\,\,\,?\,\,\,\,\,\,\,\left( {\frac{{77}}{4}=19R1\text{, so same as }{{i}^{1}}} \right)\,\,\,\,\,\,\text{So, }\color{#800000}{{{{i}^{{77}}}}}=i\\\color{#800000}{{{{i}^{{110}}}}}=\,\,\,?\,\,\,\,\,\,\left( {\frac{{110}}{4}=27R2\text{, so same as }{{i}^{2}}} \right)\,\,\,\,\,\text{So, }\color{#800000}{{{{i}^{{110}}}}}=-1\end{array}\)

Again, when dealing with complex numbers, expressions contain a **real** part and an **imaginary** part. Together they form a **complex** number that typically looks like \(a+bi\), where “\(a\)” is the real part, and “\(b\)” is the imaginary part, following by “\(i\)”, to indicate the “imaginary” unit. (Later, in **Pre-Calculus**, we’ll see how these can be graphed on a coordinate system, where the “\(x\)” is the real part and the “\(y\)” is the imaginary part.)

For example, “\(4+5i\)” indicates the number \(\displaystyle 4+\left( 5 \right)\left( {\sqrt{{-1}}} \right)\), and we **cannot mix the real parts with the imaginary parts **when adding or subtracting, so that the “\(i\)’s” are treated somewhat like variables (like radicals were thought as variables, back in the **Exponents and Radicals in Algebra** section).

So when we perform operations on \(i\), we pretty much treat it like a variable, except when we’re multiplying the “\(i\)’s” together – and then we can simplify. **Note that for good “math grammar” we want our final answer to be in the form \(\boldsymbol{a+bi}\)****.** Here are some examples:

You can also put complex expressions in the **graphing calculator**:

(Note that the **complex conjugate** that we used to simplify a denominator with an imaginary number in it is similar to the **radical conjugate** we learned about **here **in the **Introduction to Quadratics** section.)

# Quadratic Formula with Complex Solutions

Now let’s solve a quadratic equation that has complex (imaginary) solutions.

Let’s take the equation \({{x}^{2}}-2x+2\). We know that since the discriminant \(\left( {{{b}^{2}}-4ac} \right)\) for \(a,\,b,\) and \(c\) in \(a{{x}^{2}}+bx+c=0\) is negative ( \(-4\)), there are no real solutions to the equation, but there are two imaginary solutions. (Note that if there are imaginary solutions, there are always **two** of them.)

Let’s use the quadratic equation to find this solution, and one that’s a little more complicated:

# Completing the Square with Complex Solutions

Let’s try **completing the square** with a quadratic with complex solutions:

Yeah! We got the same answers as when we solved with the Quadratic Equation!

We learned earlier **here** in the **Introduction to Quadratics** section that when we have an irrational value for a root, the **conjugate** is also a root. (For example, if \(\displaystyle 3+\sqrt{{17}}\) is a root, then \(\displaystyle 3-\sqrt{{17}}\) is also a root).

Similarly, if we have a complex root, the **complex conjugate** is also a root; this is called the **Complex Conjugate Root Theorem**, or **Complex** **Conjugate Zeros Theorem**. For example, if \(3+i\)** **is a root, then \(3-i\) is also a root (the conjugate is always the second imaginary solution). Interesting!

**Learn these rules and practice, practice, practice!**

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On to **Compositions of Functions, Even and Odd, and Increasing and Decreasing** – you are ready!